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(Solved): For a square matrix A, the statement (A-lambda I)v=0hasanon-zerosolutionv is equivalent to the sta ...
For a square matrix
A, the statement
(A-lambda I)v=0hasanon-zerosolutionv is equivalent to the statement
:'det(A-\lambda I)=0 : The quantity
det(A-\lambda I)is a polynomial in
\lambda known as the characteristic polynomial of
Aand its roots are the eigenvalues of
A. This gives us a technique for finding the eigenvalues of
ALet's check our understanding. i) The characteristic polynomial of
A=([1,2],[2,1])is
det(A-tI)=det([1-t,2],[2,1-t])=(1-t)^(2)-4,0 a Ordered
t_(1)=◻t_(2)=◻◻v_(1)=◻v_(2)=◻((1)/(2))(:1,2:)B=([5,-1,0],[-1,5,0],[0,0,-3])det(B-tI)=◻Bt_(1)=-3,t_(2)=◻,t_(3)=◻ 目, t_(1) this has roots (and hence B has eigenvalues)
t_(1)=-3,t_(2)=◻,t_(3)=◻ 目, t_(1) the roots of this polynomial are t_(1)= ◻ and t_(2)=
◻ Ordered the same way, the associated eigenvectors are
◻
v_(1)=
◻
and v_(2)= ◻
Note: the Maple notation for the vector ((1)/(2)) is (:1,2:)
ii